Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Have the following let $P_2(R)$ denote a vector space of the real polynomial functions of degree less than or equal to two and let $B:=[P_0,P_1,P_2]$ denote the natural ordered basis for $P_2(R)$ (so $P_i(x)=x^{i})$ . Define $f\in P_2(R)$ by $f(x)=5x^{2}-2x+3$ . Write f as a linear combination of the elements of B. Compute the coordinate vector $f_B$ of f with repect to B.

$f:= 5p_2-2p_1+3p_0$ $\rightarrow$ $f_B:=[3,-2,5]$

Define $h_1,h_2,h_3\in P_2(R)$ by $h_1(x)= 7x^{2}+3x+4$, $h_2(x)= x^{2}+2x+1$, $h_3(x)= x^{2}-1$. Define $C:=[h_1,h_2,h_3]$. Assume C is an ordered basis for $P_2(R)$, construct the change of coordinate matrix, A, which converts C-coordinate to B-coordinates. Compute $A^{-1}$.

$A= \begin{pmatrix}4 & 1 & -1\\3 & 2 & 0\\7 &1 &1\\\end{pmatrix}$ $\Rightarrow$ $A^{-1}$= $\begin{pmatrix}2/16 & -2/16 & 2/16\\-3/16 & 11/16 & -3/16\\-11/16 &3/16 &5/16\\\end{pmatrix}$

Im struggling with the following Let F: $P_2(R)$ $\rightarrow$ $P_3(R)$ be the linear transformation determined by: $$F(f)(x)=6\int_{0}^{2x-1} f(t) dt$$

Compute the dimension of the kernel of F. Determine a basis for the image of F. Define $A:=[P_0,P_1,P_2,P_3]$ and compute $M_{B}^{A}(F)$, the matrix of F with respect to the given ordered bases. Determine the rank of $M_{C}^{A}(F)$.

Many thanks in advance.

share|cite|improve this question
What is giving you trouble? Verify that $F(f)$ is indeed a polynomial of degree at most $4$; to compute $M^A_B(F)$, evaluate $F$ at the basis vectors in $B$, and express those images in terms of $A$; you can use the matrix $A$ (bad notation alert! You are using $A$ for two different things: the basis of $P_3(R)$, and the matrix you found earlier) to compute the matrix relative to C instead of relative to B. – Arturo Magidin Feb 24 '12 at 21:23
up vote 2 down vote accepted

Note that $$6\int_0^{2x-1}(at^2+bt+c)dt=16ax^3+(12b-24a)x^2+(12a-12b+12c)x-2a+3b-6c$$ so as a linear map from $P_2(R)$ to $P_3(R)$, $F$ maps $$aP_2+bP_1+cP_0\mapsto 16aP_3+(-24a+12b)P_2+(12a-12b+12c)P_1+(-2a+3b-6c)P_0$$ which should suggest both a basis for the image and how to write the matrix $M_B^A(F)$. Given a basis for the image you can determine its dimension, and since you know the dimension of $P_2(R)$ you can apply the rank-nullity theorem to get the dimension of $\ker F$. To determine the rank of $M_C^A(F)$, recall that it is the same transformation as $M_B^A(F)$, just with respect to a different basis. What does this say about the ranks of $M_C^A(F)$ and $M_B^A(F)$?

share|cite|improve this answer
Think I may be getting a little ahead of myself in the syllabus, trying to get a bit ahead but am a little lost. – user24930 Feb 25 '12 at 15:30
I know its been a long time since this post, just concerning the last bit, would the ranks be equal? – user24930 Apr 1 '12 at 10:00
@user24930 Yes. – Alex Becker Apr 1 '12 at 14:33

Assuming you know how to determine the basis of the kernel of a linear transformation given by a matrix, the first problem can be solved by finding the matrix of the transformation $F$. That again is a matter of figuring out what $F$ does to basis vectors; consider a linear transformation $F: V \to W$ and let $\{u_i\}_{i = 1}^{m} \subseteq V$ be a basis for the domain $V$ and let $\{v_j\}_{j = 1}^{n} \subseteq W$ be a basis for the codomain $W$. To determine the matrix element $f_{ij}$ in the matrix $F = [f_{ij}]$, take the $i$'th basis vector $u_i$ in your chosen basis for the domain, apply the transformation $F$ and write the result in terms of the basis for the codomain. The coefficient $f_{ij}$ is simply the constant in front of the $j$'the basis vector, i.e. $$Fu_i = \sum_{j = 1}^{n} f_{ij} v_j $$  

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.